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The dual gauge field, $V$, is defined by $$^{\star}F(V)=F(A),$$ where $F$ is the field strength. The 't Hooft operator $\exp(i\int_C V)$ creates the trajectory of a magnetic particle along $C$. But I don't really understand what this has to do with topology? There are two topological quantities that I am aware of:

  • The topological number of a four-dimensional configuration which is a pure gauge at infinity ($S^3_{\infty}$) is $$\int d^4x\,\operatorname{tr}(F \tilde F).$$
  • The winding number of a gauge transformation $\Omega(\mathbf x)$ is $$\int d^3x\,\epsilon^{ijk}\operatorname{tr}(\Omega^{-1}\partial_i \Omega~ \Omega^{-1}\partial_j \Omega ~\Omega^{-1}\partial_k \Omega).$$

However, neither of these allows me to compute the topological charge of a magnetic particle. So my question is: How can I conclude that the magnetic particle is a topological excitation?

DeltaTron
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1 Answers1

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The 't Hooft operator creates a magnetic field which is a delta function smeared along the curve $C$. In this question here it was shown that a gauge transformation which is singular (i.e. multivalued) on a curve $C$ produces a magnetic field that is a delta function smeared along $C$. Therefore, the effect of the 't Hooft operator is to act with a singular gauge transformation of this kind.

But such gauge transformations are classified by $\mathbb Z^{N-1}$ ($N$ being the number of colors), which counts the winding of the $(N-1)$ $U(1)$'s of the Cartan subgroup around the curve $C$. So such a singular gauge transformation is said to create a magnetic monopole with charge $m\in\mathbb Z^{N-1}$.

DeltaTron
  • 800